In the previous section, we saw how to calculate εsc. Now we will use it to calculate fsc, the stress in Asc. We know that stress = strain x Es. Where Es is the young's modulus of steel. For a given strain, we can calculate the corresponding stress from this relation. If the steel is of Fe250 grade, the calculation of stress is easy, since the stress varies linearly with strain (Fig.23B of the code). But for cold formed bars like Fe 415 and Fe 500, the stress calculation is easy only if the strain falls in the straight line portions. For example, in the case of Fe 415 steel (Fig.3.13), The stress can be easily calculated if the strain falls in the initial red colored inclined line or the final magenta colored horizontal line. If the strain falls in the blue colored curved portion, we have to use linear interpolation between appropriate values. We have obtained the tables which give the values to be used for this interpolation.
For convenience, the tables are shown again below:
The above method for calculating fsc involves lengthy calculations. We will discuss another easier method:
In the previous section, we have derived Eq.11.1 for calculating εsc.
Eq.11.1
In this equation, we will bring in the effective depth d.
For Fe415 steel, xumax / d is equal to 0.4791 (Table.3.4). So xumax = 0.4791d . Substituting this in 11.1, we get
Eq.11.2
Similarly, for Fe500 steel, we get
Eq.11.3
So we get εsc in terms of d'/d ratio. In practice the ratio d'/d is found to vary in the range 0.05 to 0.20 .We can think of four values 0.05, 0.10, 0.15 and 0.20, and plugin these four values in Eqs.11.2 and 11.3. So each of those equations will give four values for εsc . For each of these values of εsc, we can calculate fsc from the above tables for Fe415 and Fe500. The results can be written in a tabular form as shown below. fsc corresponding to intermediate values of d'/d can be calculated by linear interpolation.
Table 11.1
Sample calculation 1:
Let us take Fe415, and a d'/d ratio of 0.15
Plugin 0.15 into the equation 11.2. We get εsc = 0.0024042
Now we take the appropriate values from the table for Fe 415 steel, given above.
strain = 0.0019200, stress = 324.80
strain = 0.0024042, stress = 342.59
strain = 0.0024100, stress = 342.80
Using linear interpolation, the value of stress for a strain of 0.0024042 is equal to 342.59 N/mm2
Sample calculation 2:
Let us take Fe500, and a d'/d ratio of 0.05
Plugin 0.05 into the equation 11.6. We get = 0.0031162
Now we take the appropriate values from the table for Fe500 steel, given above.
strain = 0.0027700, stress = 413.00
strain = 0.0031162, stress = 423.78
strain = 0.0031200, stress = 423.90
Using linear interpolation, the value of stress for a strain of 0.0031162 is equal to 423.78 N/mm2
Sample calculation 3:
Here we look at the case of Fe250 steel. We see that for all the d'/d ratios, the value of stress in table 11.1 above, is equal to 217.50 N/mm2. We will now look into the reason for this. Fig. 23B of the code shows the stress strain relation ship of Fe250 steel. We have already discussed about it, and we know that, when the stress reaches a value of 0.87fy, Fe250 will yield, and will continue to undergo strain at constant stress. This yielding at constant stress is shown by the horizontal line. If we draw a vertical line at the beginning of this horizontal line, the point of intersection of it with the X axis will give the strain at yield point. In our case, the strains corresponding to all the d'/d ratios (from 0.05 to 0.20) are greater than the strain at yield point. So stress corresponding to these strains will all fall in the horizontal portion of the graph, as shown in the fig. below. So we get all the values same.
Fig.11.3
Strain values when d'/d ratio is in the range 0.05 - 0.20
This completes the discussion about the above table 11.1. With this table, we can easily calculate fsc. The product of this stress fsc and area of steel Asc will give the force in that steel. But we have to note one point here. When the steel of area Asc is placed in the section, some concrete will be removed. The area of the concrete which is removed will be equal to Asc. We will not receive any contribution from this removed concrete area. This removal is not considered in the calculation of Mulim. There we take the whole concrete area above the NA. So we will make the required deduction now:
• The area of concrete = Asc.
• Stress in concrete = 0.447fck. This is because the top portion of the beam section will experience a constant stress of 0.447fck, as indicated by the rectangular portion of the stress block. This 'constant' nature of the stress will exist for a depth, equal to the depth of the rectangular portion, which is equal to 3/7xu.
So force = 0.447fck Asc. This much should be deducted. Thus the net force Cus = Asc fsc - 0.447fck Asc = (fsc - 0.447fck)Asc
This is the extra new force which gives the extra moment capacity of ΔMu. But this new force should be balanced by another force, which is tensile in nature. For this we provide some tensile steel (in addition to the already present Ast,lim) at the bottom portion of the beam. We denote this extra tensile steel as ΔAst.
This extra tensile steel is provided in the same layer as Ast,lim. So just as for Ast,lim, the c.g of ΔAst is also at a distance of d from the top most compression fibre. If the bars are provided in more than one layers, the c.g of the whole bar group (which comprises of both Ast,lim and ΔAst) is considered. So in the case of multiple layers also, the c.g of the ΔAst is at a distance of d from the top most compression fibre.
Next we will calculate the force in this new tensile steel ΔAst. We know that, at the ultimate state, the tensile steel should have yielded. This should be checked for every newly designed beam, whether the tensile steel is provided in a single layer or multiple layers. What ever be the number of layers, at the ultimate state, the strain at the c.g of the bar group should be greater than or equal to the 'yield strain'. So the stress in ΔAst will also be equal to 0.87fy. Thus the force ΔTu = 0.87fy ΔAst.
We now know the two new forces: Cus and ΔTu . The distance between them is d-d'. So the extra moment capacity ΔMu = force x lever arm. We can use either the tensile force ΔTu or the compressive force Cus.
Thus we can write:
Eq.11.4
ΔMu = (fsc - 0.447fck)Asc (d - d'). and
Eq.11.5
ΔMu = 0.87fy ΔAst (d - d')
In the next section we will see the application of these two equations.
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For convenience, the tables are shown again below:
Table for Fe415:
| Strain | Stress (N/mm2) | Calculation of Strain | Calculation of Stress |
| 0.000 | 0.00 | ||
| 0.00144 | 288.7 | (415x0.80)/230000 | (415 x 0.80)/1.15 |
| 0.00163 | 306.7 | 0.0001+[(415x0.85)/230000] | (415 x 0.85)/1.15 |
| 0.00192 | 324.8 | 0.0003+[(415x0.90)/230000] | (415 x 0.90)/1.15 |
| 0.00241 | 342.8 | 0.0007+[(415x0.95)/230000] | (415 x 0.95)/1.15 |
| 0.00276 | 351.8 | 0.001+[(415x0.975)/230000] | (415 x 0.975)/1.15 |
| ≥ 0.00380 | 360.9 | 0.002+[(415x1.0)/230000] | (415 x 1.00)/1.15 |
Table for Fe500:
| Strain | Stress (N/mm2) | Calculation of Strain | Calculation of Stress |
| 0.000 | 0.00 | ||
| 0.00174 | 347.8 | (500x0.80)/230000 | (500 x 0.80)/1.15 |
| 0.00195 | 369.6 | 0.0001+[(500x0.85)/230000] | (500 x 0.85)/1.15 |
| 0.00226 | 391.3 | 0.0003+[(500x0.90)/230000] | (500 x 0.90)/1.15 |
| 0.00277 | 413.0 | 0.0007+[(500x0.95)/230000] | (500 x 0.95)/1.15 |
| 0.00312 | 423.9 | 0.001+[(500x0.975)/230000] | (500 x 0.975)/1.15 |
| ≥ 0.00417 | 434.8 | 0.002+[(500x1.0)/230000] | (500 x 1.00)/1.15 |
In the previous section, we have derived Eq.11.1 for calculating εsc.
Eq.11.1
In this equation, we will bring in the effective depth d.
For Fe415 steel, xumax / d is equal to 0.4791 (Table.3.4). So xumax = 0.4791d . Substituting this in 11.1, we get
Eq.11.2
Similarly, for Fe500 steel, we get
Eq.11.3
So we get εsc in terms of d'/d ratio. In practice the ratio d'/d is found to vary in the range 0.05 to 0.20 .We can think of four values 0.05, 0.10, 0.15 and 0.20, and plugin these four values in Eqs.11.2 and 11.3. So each of those equations will give four values for εsc . For each of these values of εsc, we can calculate fsc from the above tables for Fe415 and Fe500. The results can be written in a tabular form as shown below. fsc corresponding to intermediate values of d'/d can be calculated by linear interpolation.
Table 11.1
Sample calculation 1:
Let us take Fe415, and a d'/d ratio of 0.15
Plugin 0.15 into the equation 11.2. We get εsc = 0.0024042
Now we take the appropriate values from the table for Fe 415 steel, given above.
strain = 0.0019200, stress = 324.80
strain = 0.0024042, stress = 342.59
strain = 0.0024100, stress = 342.80
Using linear interpolation, the value of stress for a strain of 0.0024042 is equal to 342.59 N/mm2
Sample calculation 2:
Let us take Fe500, and a d'/d ratio of 0.05
Plugin 0.05 into the equation 11.6. We get = 0.0031162
Now we take the appropriate values from the table for Fe500 steel, given above.
strain = 0.0027700, stress = 413.00
strain = 0.0031162, stress = 423.78
strain = 0.0031200, stress = 423.90
Using linear interpolation, the value of stress for a strain of 0.0031162 is equal to 423.78 N/mm2
Sample calculation 3:
Here we look at the case of Fe250 steel. We see that for all the d'/d ratios, the value of stress in table 11.1 above, is equal to 217.50 N/mm2. We will now look into the reason for this. Fig. 23B of the code shows the stress strain relation ship of Fe250 steel. We have already discussed about it, and we know that, when the stress reaches a value of 0.87fy, Fe250 will yield, and will continue to undergo strain at constant stress. This yielding at constant stress is shown by the horizontal line. If we draw a vertical line at the beginning of this horizontal line, the point of intersection of it with the X axis will give the strain at yield point. In our case, the strains corresponding to all the d'/d ratios (from 0.05 to 0.20) are greater than the strain at yield point. So stress corresponding to these strains will all fall in the horizontal portion of the graph, as shown in the fig. below. So we get all the values same.
Fig.11.3
Strain values when d'/d ratio is in the range 0.05 - 0.20
This completes the discussion about the above table 11.1. With this table, we can easily calculate fsc. The product of this stress fsc and area of steel Asc will give the force in that steel. But we have to note one point here. When the steel of area Asc is placed in the section, some concrete will be removed. The area of the concrete which is removed will be equal to Asc. We will not receive any contribution from this removed concrete area. This removal is not considered in the calculation of Mulim. There we take the whole concrete area above the NA. So we will make the required deduction now:
• The area of concrete = Asc.
• Stress in concrete = 0.447fck. This is because the top portion of the beam section will experience a constant stress of 0.447fck, as indicated by the rectangular portion of the stress block. This 'constant' nature of the stress will exist for a depth, equal to the depth of the rectangular portion, which is equal to 3/7xu.
So force = 0.447fck Asc. This much should be deducted. Thus the net force Cus = Asc fsc - 0.447fck Asc = (fsc - 0.447fck)Asc
This is the extra new force which gives the extra moment capacity of ΔMu. But this new force should be balanced by another force, which is tensile in nature. For this we provide some tensile steel (in addition to the already present Ast,lim) at the bottom portion of the beam. We denote this extra tensile steel as ΔAst.
This extra tensile steel is provided in the same layer as Ast,lim. So just as for Ast,lim, the c.g of ΔAst is also at a distance of d from the top most compression fibre. If the bars are provided in more than one layers, the c.g of the whole bar group (which comprises of both Ast,lim and ΔAst) is considered. So in the case of multiple layers also, the c.g of the ΔAst is at a distance of d from the top most compression fibre.
Next we will calculate the force in this new tensile steel ΔAst. We know that, at the ultimate state, the tensile steel should have yielded. This should be checked for every newly designed beam, whether the tensile steel is provided in a single layer or multiple layers. What ever be the number of layers, at the ultimate state, the strain at the c.g of the bar group should be greater than or equal to the 'yield strain'. So the stress in ΔAst will also be equal to 0.87fy. Thus the force ΔTu = 0.87fy ΔAst.
We now know the two new forces: Cus and ΔTu . The distance between them is d-d'. So the extra moment capacity ΔMu = force x lever arm. We can use either the tensile force ΔTu or the compressive force Cus.
Thus we can write:
Eq.11.4
ΔMu = (fsc - 0.447fck)Asc (d - d'). and
Eq.11.5
ΔMu = 0.87fy ΔAst (d - d')
In the next section we will see the application of these two equations.
PREVIOUS
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